《物理双语教学课件》Chapter 9 Oscillations 振动（合集）

第一篇：《物理双语教学课件》Chapter 9 Oscillations 振动

Chapter 9 Oscillations

We are surrounded by oscillations─motions that repeat themselves.(1).There are swinging chandeliers, boats bobbing at anchor, and the surging pistons in the engines of cars.(2).There are oscillating guitar strings, drums, bells, diaphragms in telephones and speaker systems, and quartz crystals in wristwatches.(3).Less evident are the oscillations of the air molecules that transmit the sensation of sound, the oscillations of the atoms in a solid that convey the sensation of temperature, and the oscillations of the electrons in the antennas of radio and TV transmitters.Oscillations are not confined to material objects such as violin strings and electrons.Light, radio waves, x-rays, and gamma rays are also oscillatory phenomena.You will study such oscillations in later chapters and will be helped greatly there by analogy with the mechanical oscillations that are about to study here.Oscillations in the real world are usually damped;that is, the motion dies out gradually, transferring mechanical energy to thermal energy by the action of frictional force.Although we cannot totally eliminate such loss of mechanical energy, we can replenish the energy from some source.9.1 Simple Harmonic Motion 1.The figure shows a sequence of “snapshots” of a simple oscillating system, a particle moving repeatedly back and forth about the origin of the x axis.2.Frequency:(1).One important property of oscillatory motion is its frequency, or number of oscillations

that

are completed each second.(2).The symbol for frequency is f, and(3)its SI unit is hertz(abbreviated Hz), where 1 hertz = 1 Hz = 1 oscillation per second = 1 s-1.3.Period: Related to the frequency is the period T of the motion, which is the time for one complete oscillation(or cycle).That is T1f.4.Any motion that repeats itself at regular intervals is called period motion or harmonic motion.We are interested here in motion that repeats itself in a particular way.It turns out that for such motion the displacement x of the particle from the origin is given as a function of time by

x(t)xmcos(t), in which xm,,and

are constant.The motion is called simple harmonic motion(SHM), the term that means that the periodic motion is a sinusoidal of time.5.The quantity

xm, a positive constant whose value depends on how the motion was started, is called the amplitude of the motion;the subscript m stands for maximum displacement of the particle in either direction.6.The time-varying quantity

(t)

is called the phase of the motion, and the constant  is called the phase constant(or phase angle).The value of  depends on the displacement and velocity of the particle at t=0.7.It remains to interpret the constant .The displacement

x(t)

must return to its initial value after one period T of the motion.That is，x(t)

must equal

0x(tT)for all t.To simplify our analysis, we put.So we then have xmcostxmcos[(tT)].The cosine function first repeats itself when its argument(the phase)has increased by that we must have 22fT2 rad, so

(tT)t2orT2.It means.The quantity  is called the angular frequency

of the motion;its SI unit is the radian per second.8.The velocity of SHM:(1).Take derivative of the displacement with time, we can find an expression for the velocity of the particle moving with simple harmonic motion.That is, v(t)dx(t)xmsin(t)vmcos(t/2).(2).The dtpositive quantity

vmxm in above equation is called the velocity amplitude.9.The acceleration of SHM: Knowing the velocity for simple harmonic motion, we can find an expression for the acceleration of the oscillation particle by differentiating once more.Thus we have

a(t)dv(t)vmsin(t/2)amcos(t)dtThe positive quantity

amvm2xm is called the acceleration

a(t)2x(t)amplitude.We can also to get , which is the hallmark of simple harmonic motion: the acceleration is proportional to the displacement but opposite in sign, and the two quantities are related by the square of the angular frequency.9.2 The Force Law For SHM 1.Once we know how the acceleration of a particle varies with time, we can use Newton’s second law to learn what force must act on the particle to give it that acceleration.For simple harmonic motion, we have

Fma(m2)xkx.This result-a force proportional to the displacement but opposite in sign-is something like Hook’s law for a spring, the spring constant here being km2.2.We can in fact take above equation as an alternative definition of simple harmonic motion.It says: Simple harmonic motion is the motion executed by a particle of mass m subject to a force that is proportional to the displacement of the particle but opposite in sign.3.The block-spring system forms a linear simple harmonic oscillator

(linear oscillator for short), where linear indicates that F is proportional to x rather than to some other power of x.(1).The angular frequency  of the simple harmonic motion of the block is oscillator is

9.3 Energy in Simple Harmonic Motion 1.The potential energy of a linear oscillator depends on how much the spring is stretched or compressed, that is, on

k/m.(2).The period of the linear T2m/k.x(t).We have U(t)1212kxkxmcos2(t).222.The kinetic energy of the system depends on haw fast the block is moving, that is on

K(t)v(t).We have 1212mvm2xmsin2(t)22

1k212m()xmsin2(t)kxmsin2(t)2m23.The mechanical energy is

EUK121212kxmcos2(t)kxmsin2(t)kxm 222The mechanical energy of a linear oscillator is indeed a constant, independent of time.9.4 An Angular simple Harmonic Oscillator 1.The figure shows an angular version of a simple harmonic oscillator;the element of springiness or elasticity is associated with the twisting of a suspension wire rather than the extension and compression of a spring as we previously had.The device is called a torsion pendulum, with torsion referring to the twisting.2.If we rotate the disk in the figure from its rest position and release it, it will oscillate about that position in angular simple harmonic motion.Rotating the disk through an angle  in either direction introduce a restoring torque given by Here (Greek kappa)is a constant, called the .torsion constant, that depends on the length, diameter, and material of the suspension wire.3.From the parallelism between angular quantities and linear quantities(give a little more explanation), we have

T2I

for the period of the angular simple harmonic oscillator, or torsion pendulum.9.5 Pendulum We turn now to a class of simple harmonic oscillators in which the springiness is associated with the gravitational force rather than with the elastic properties of a twisted wire or a compressed or stretched spring.1.The Simple Pendulum(1).We consider a simple pendulum, which consists of a particle of mass m(called the bob of the pendulum)suspended from an un-stretchable, massless string of length L, as in the figure.The bob is free to swing back and forth in the plane of the page, to the left and right of a vertical line through the point at which the upper end of the string is fixed.(2).The forces acting the particle, shown in figure(b), are its weight and the tension in the string.The restoring force is the tangent component of the weight

mgsin, which is always acts opposite the displacement of the particle so as to bring the particle back toward its central location, the equilibrium(0).We write the restoring force as Fmgsin, where the minus sign indicates that F acts opposite the displacement.(3).If we assume that the angle is small, the

sin is very nearly equal to  in radians, and the displacement s of the particle measured along its arc is equal to FmgmgL.Thus, we have

smg()s.Thus if a simple pendulum swings LLthrough a small angle, it is a linear oscillator like the block-spring oscillator.(4).Now the amplitude of the motion is measure as the angular amplitude m, the maximum angle of swing.The period of

a

simple

pendulum

is T2m/k2m/(mg/L)2L/g.This result hods only if the angular amplitude m is small.2.The Physical Pendulum

(1).The figure shows a generalized physical pendulum, as we shall call realistic pendulum, with its weight mg

acting at the center of mass C.(2).When the pendulum is displaced through an angle  in either direction from its equilibrium position, a restoring torque appears.This torque acts about an axis through the suspension point O in the figure and has the magnitude (mgsin)(h).The minus sign indicates that the torque is a restoring torque, which always acts to reduce the angle  to zero.(3).We once more decide to limit our interest to small amplitude, so that (mgh).sin.Then the torque becomes

T2I/mgh,(4).Thus the period of a physical pendulum is when m is small.Here I is the rotational inertia of the pendulum.(5).Corresponding to any physical pendulum that oscillates about a given suspension point O with period T is a simple pendulum of length L0 with the same period T.The point along the physical pendulum at distance L0 from point O is called the center of oscillation of the physical pendulum for the given suspension point.3.Measuring g: We can use a physical pendulum to measure the free-fall acceleration g through measuring the period of the pendulum.9.6 Simple Harmonic Motion and Uniform circular Motion 1.Simple harmonic motion is the projection of uniform circular motion on a diameter of the circle in which the latter motion occurs.2.The figure(a)gives an example.It shows a reference moving circular particle in

P’

uniform with motion angular speed  in a reference circle.The radius xm of the circle is the magnitude of the particle’s position vector.At any time t, the angular position of the particle is t.3.The projection of particle P’ onto the x axis is a point P.The projection of the position vector of particle P’ onto the x axis gives the location x(t)of P.Thus we find

x(t)xmcos(t).Thus if reference particle P’ moves in uniform circular motion, its projection particle P moves in simple harmonic motion.4.The figure(b)shows the velocity of the reference particle.The magnitude of the velocity is xm, and its projection on the x axis is

v(t)xmsin(t).The minus sign appears because the velocity component of P points to the left, in the direction of decreasing x.5.The figure(c)shows the acceleration of the reference particle.The magnitude of the acceleration vector is projection on the x axis is

a(t)2xmcos(t).2xm

and its 6.Thus whether we look at the displacement, the velocity, or the acceleration, the projection of uniform circular motion is indeed simple harmonic motion.9.7 Damped Simple Harmonic Motion A pendulum will swing hardly at all under water, because the water exerts a drag force on the pendulum that quickly eliminates the motion.A pendulum swinging in air does better, but still the motion dies out because the air exerts a drag force on the pendulum, transferring energy from the pendulum’s motion.1.When the motion of an oscillator is reduced by an external force, the oscillator and its motion are said to be damped.An idealized example of a damped oscillator is shown in the figure: a block with mass m oscillates on a spring with spring constant k.From the mass, a rod extends to a vane(both assumed massless)that is submerged in a liquid.As the vane moves up and down, the liquid exerts an inhibiting drag force on it and thus on the entire oscillating system.With time, the mechanical energy of the block-spring system decreases, as energy is transferred to thermal energy of the liquid and vane.2.Let us assume that the liquid exerts a damped forceFd that is

proportional in magnitude to the velocity v of the vane and

block.Then

Fdbv，where b is a damped constant that depends on the characteristics of both the vane and the liquid and has the SI unit of kilogram per second.The minus sign indicates that

Fd

opposes the motion.3.The total force acting on the block is Fkxbvkxbdx.dtSo we have equation

d2xdxm2bkx0，dtdtwhose solution is x(t)xmebt/2mcos('t), where ', the angular frequency of the 12 damped oscillator, is given by 'kb2m4m2.4.We can regard the displacement of the damped oscillator as a cosine function whose amplitude, which is decreases with time.5.The mechanical energy of a damped oscillator is not constant but decreases with time.If the damping is small, we can find E(t)

xmebt/2m, gradually by replacing

xm

with

xmebt/2m，the amplitude of the

E(t)12bt/mkxme, which 2damped oscillation.Doing so, we find tells us that the mechanical energy decreases exponentially with time.9.8 Forced Oscillations and Resonance 1.A person swing passively in a swing is an example of free oscillation.If a kind friend pulls or pushes the swing periodically, as in the figure, we have forced, or driven, oscillations.There are now two angular frequencies with which to deal with:(1)the natural angular frequency  of the system, which is the angular frequency at which it would oscillate if it were suddenly disturbed and then left to oscillate freely, and(2)the angular frequency d of the external driving force.2.We can use the right figure to represent an idealized forced simple harmonic oscillator if we allow the structure marked “rigid support” to move up and down at a variable angular frequency d.A forced oscillator oscillates at the angular frequency d of driving force, and its displacement is given by

x(t)xmcos(dt), where xm

is the amplitude of the oscillations.How large the displacement amplitude and .3.The velocity amplitude

vm xm

is depends on a complicated function of d

of the oscillations is easier to describe: it is greatest when d, a condition called resonance.Above equation is also approximately the condition at which the displacement amplitude

xm

of oscillations is greatest.The figure shows how the displacement amplitude of an oscillator depends on the angular frequency

d of the driving force, for three values of the damped coefficient b.4.All mechanical structures have one or more natural frequencies, and if a structure is subjected to a strong external driving force that matches one of these frequencies, the resulting oscillations of structure may rupture it.Thus, for example, aircraft designers must make sure that none of the natural frequencies at which a wing can vibrate matches the angular frequency of the engines at cruising speed.15

第二篇：神经解剖学双语教学课件

神经解剖学双语教学课件

1． 感觉器官Sensory organs.ppt

2． 神经系统总论、脊髓Spinal cord.ppt

3． 脑干Brain stem.ppt

4． 小脑、间脑Cerebellum.ppt

5． 端脑

6． 神经传导通路

7． 脑和脊髓的被膜和血管

8． 脊神经

9． 脑神经 10．

11． 内脏神经、内分泌系统 帕金森病

第三篇：《物理双语教学课件》Chapter 15 Electric Fields 电场

Chapter 15 Electric Fields

Suppose we fix a positively charged particle q1 in place and then put a second positively charged particle q2 near it.From Coulomb's law we know that q1 exerts a repulsive electrostatic force on q2.Then you may ask how q1 “know” of the presence of q2? That is, since the charges do not touch, how can q1 exert a force on q2? This question about action at a distance can be answered by saying that q1 set up an electric field in the space surrounding it.At any given point P in that space, the field has both magnitude and direction.Thus when we place q2 at P, q1 interacts with q2 through the electric field at P.The magnitude and direction of that electric field determine the magnitude and direction of the force acting on q2.Another action-at-a-distance problem arises if we move q1, say, toward q2.Coulomb’s law tells us that when q1 is closer to q2, the repulsive electrostatic force acting on q2 must be greater, and it is.Does the electric field at q2, and thus the force acting on q2, change immediately? The answer is no.Instead, the information about the move by q1 travels outward from q1 as electromagnetic wave at the speed of light c.The change in the electric field at q2, and thus the change in the force acting on q2, occurs when the wave finally reach q2.15.1 The Electric Field 1.The temperature at every point in a room has a definite value.We call the resulting distribution of temperature as temperature field.In much the same way, you can imagine a pressure field in the atmosphere: it consists of the distribution of air pressure values, one for each point in the atmosphere.Theses two examples are of scalar field, because temperature and air-pressure are scalar quantities.2.The electric field is a vector field

(1)It consists of a distribution of vectors, one for each point in the region around a charged object.(2)In principle, we can define the electric field at some point near the charged object by placing a positive charge q0, called a test charge, at the point.(3)We then measure the electrostatic force test charge.The electric field object is defined as

FEq0E

F

that acts on the

at point P due to the charged

.(4)We represent the electric field at point P with a vector whose tail is at P, as shown in the figure.(5)The SI unit for the electric field is the newton per coulomb(N/C).15.2 Electric Field Lines 1.Michael Fraday, who introduced the idea of electric fields in the 19th century, thought of the space around a charged body as filled with lines of force.Although we no longer attach much reality to these lines, now usually called electric field lines, they still provide a nice way to visualize patters in electric fields.2.The relation between the field lines and electric field vectors(1)At any point, the direction of a straight field line or the direction of the tangent to a curve field line gives the direction of E at that point.(2)The field lines are drawn so that the number of lines per unit area, measured in a plane that is perpendicular to the lines, is proportional to the magnitude of

E.This second relation means that where the field lines are close together, E is large;and where they are far apart, E is small.3.Some Electric Field lines(1)The electric field lines of a sphere with uniform charge as shown in the right figure.(2)Right figure gives the electric field lines of an infinitely large, non-conducting sheet(or plane)with a uniform distribution of positive charge on one side.(3)The figure shows the field lines for two equal positive point charges.(4)The figure shows the pattern for two charges that are equal in magnitude but opposite sign.(5)From above figures, we can come to the conclusion: Electric field lines extend away from positive charge and toward negative charge.15.3 The Electric Fields for some cases 1.The electric field due to a point charge q

(1)If we put a positive test charge q0 at any point a distance r from the point charge, the magnitude of the electrostatic force acting on q0, from Coulomb’s law, is magnitude of the electric field vector is direction of E

F1qq0240r.The.The

EF1qq040r2is the same as that of the force on the positive test charge: directly away from the point charge, as shown in right figure, if q is positive, and toward it if q is negative.(2)We can find the net, or resultant, electric field due to more than one point charges with the aid of the principle of superposition.If we place a positive test charge q0 near n point charges q1, q2,…,qn, then the net force charges acting on the test charge is

F0

from the n point

F0F01F02F0n.So the net electric field at the position of the test charge is F0F01F02F0nEE1E2En.Here Ei q0q0q0q0is the electric field that would be set up by point charge i acting alone.5 2.The electric field due to an electric dipole: Figure(a)shows two

charges

of magnitude q but of opposite sign, separated by a distance d.We call this configuration an electric dipole.Let us find the electric field due to the dipole at a point P, a distance z from the midpoint of the dipole and on its central axis, as shown in the figure.(1)The magnitude of the electric field is Eqd1p20z320z31, in which the product qd is the magnitude

p p of a vector quantity known as the electric dipole moment of the dipole.3.The electric field due to a line of charge

(1)So far we have considered the electric field that is produced by one or, at most, a few point charges.We now consider charge distributions that consist of great many closed spaced point charges(perhaps billions)that are spread along a line, over a surface, or within a volume.Such distributions are said to be continuous rather than discrete.When we deal with continuous charge distributions, it is most convenient to express the charge on an object as a charge density rather than as a total charge.For a line of charge, for example, we would report the linear charge density(or charge per length)whose SI unit is the coulomb per meter.(2)The figure shows a thin ring of radius R with a uniform positive linear charge density , around its circumference.What is the electric field at point P, a distance z from the plane of the ring along its central axis?(3)We can get the magnitude of the electric field as Eqz40(z2R2)3/2.4.The electric field due to a charged disk(1)The figure shows a circular plastic disk of radius R that has a positive surface charge uniform density  on its upper surface.What is the electric field at point P, a distance z from the disk along its central axis?

(2)Our plan is to divide the disk into concentric flat rings and then to calculate the electric field at point P by adding up the contributions of all rings.The magnitude of the electric field is Ez(1).2220zR

15.4 A point charge in an electric field 1.We now want to determine what happens to a charged particle that is in an electric field that produced by other stationary or slowly moving charges.This force is given by which q is the charge of the particle and

E

FqE, in

is the electric field that other charges have produced at the location of the particle.2.Measuring the elementary charge: Figure shows the Millikan

oil-drop apparatus for measuring the elementary charge e.3.Ink-jet printing: Figure shows the

essential features of an ink-jet printer.15.5 A Dipole in an Electric Field 1.The figure shows why the molecule of water(H2O)is an electric dipole.If the water molecule is placed in an external electric field, it behaves as would be expected of the more abstract electric dipole.2.The torque can be written as

pE.3.Potential energy of an electric dipole: The expression for the potential energy of an electric dipole in an external electric field is simplest if we choose the potential energy to be zero when the angle  in the figure is 900.It can be written as

UpE.Microwave cooking: the explanation it.

第四篇：《物理双语教学课件》Chapter 23 Interference 干涉理论

Chapter 23 Interference

Sunlight, as the rainbow shows us, is a composite of all the colors of the visible spectrum.Soup bubbles and oil slicks can also show striking colors, produced not by refraction but by constructive and destructive interference of light.23.1 Young’s Interference Experiment 1.In 1801 Thomas Young experimentally proved that light is a wave, contrary what most other scientists then thought.He did so by demonstrating that light undergoes interference.2.Figure gives the basic arrangement of Young’s double-slit interference experiment.On

the viewing screen C, Points of interference maxima form visible bright rows-called bright bands, bright fringes, or maxima.Dark regions-called dark bands, dark fringes, or minima-result from fully destructive interference and are visible between adjacent pairs of bright fringes.The pattern of bright and dark fringes on the screen is called an interference pattern.Figure is a photograph of the interference pattern.3.To find

what

exactly determines the locations of the fringes in Young’s double-slit interference experiment, let us see the

figure.The

L

r2path-length-difference between rays be written as

r1 and can Ldsin, where d is the separation of the two slits.(1)For a bright fringe, number of wavelengths.It is dark fringe, It is L

L

must be zero or an integer

dsinmm0,1,2,.(2)For a

must be an odd multiple of half a wavelength.2m1,2,.(3)Using above two equations, dsin(2m1)we can find the angle  to any fringe and thus locate that fringe;further, we can use the values of m to label the fringes.4.We now wish to derive an expression for the intensity I of the fringes as a function of .(1)Let us assume that the electric field components of the light waves arriving at point P in the figure from the two slits are not in phase and vary with time as

E1E0sint and

E2E0sin(t2dsin)respectively.(2)So we have where 2dsin.(3)Thus

11EE1E22E0cos()sin(kx)，22the intensity is I(2E0cos1)2

214I0cos2.2

5.Combining more than two waves:

23.2 Interference from Thin Films 1.The colors we see when sunlight illuminates a soap bubble or an oil slick are caused by the interference of light waves reflected from the front and back surfaces of a thin transparent film.The thickness of the soap or oil film is typically the order of magnitude of the wavelength of the light involved.2.Figure shows a thin transparent film of uniform thickness L and index of refraction n2, illuminated by bright light of wavelength  from a distant point source.For now, we assume that air lies on both sides of the film and thus that

n1n3

in figure.For simplicity, we also assume that the light rays are almost perpendicular to the film.3.Reflection phase shifts: Refraction at an interface never causes a phase change.But reflection can, depending on the indices of refraction on two sides of the interface.When an incident wave travels in the medium of lesser index of refraction(with greater speed), the wave that is reflected at the interface undergoes a phase shift of  rad, or half a wavelength.4.The optic path-length difference in the case of thin film is 2nL2.2nL2mm1,2,.5.For a bright film, we have 6.For a dark film, we have

2nL2(2m1)2m0,1,2,.23.3 Michelson’s Interferometer 1.An interferometer is a device that can be used to measure lengths or changes in length with great accuracy by means of interference fringes.2.Figure shows

Michelson’s interferometer.

第五篇：《物理双语教学课件》Chapter 6 Rotation 定轴转动

Chapter 6 Rotation In this chapter, we deal with the rotation of a rigid body about a fixed axis.The first of these restrictions means that we shall not examine the rotation of such objects as the Sun, because the Sun-a ball of gas-is not a rigid body.Our second restriction rules out objects like a bowling ball rolling down a bowling lane.Such a ball is in rolling motion, rotating about a moving axis.6.1 The Rotational Variables 1.Translation and Rotation: The motion is the one of pure translation, if the line connecting any two points in the object is always parallel with each other during its motion.Otherwise, the motion is that of rotation.Rotation is the motion of wheels, gears, motors, the hand of clocks, the rotors of jet engines, and the blades of helicopters.2.The nature of pure rotation: The right figure shows a rigid body of arbitrary shape in pure rotation around a fixed axis, called the axis of rotation or the rotation axis.(1).Every point of the body moves in a circle whose center lies on the axis of the rotation.(2).Every point moves through the same angle during a particular time interval.3.Angular position: The above figure shows a reference line, fixed in the body, perpendicular to the axis, and rotating with the body.We can describe the motion of the rotating body by specifying the angular position of this line, that is, the angle of the line relative to a fixed direction.In the right figure, the angular position

sr is measured relative to the positive direction of the x axis, and  is given by

(radianmeasure).Here s is the length of the arc(or the arc distance)along a circle and between the x axis and the reference line, and r is a radius of that circle.An angle defined in this way is measured in radians(rad)rather than in revolutions(rev)or degree.They have relations

1rev360o2r2rad r4.If the body rotates about the rotation axis as in the right figure, changing the angular position of the reference line from 1 to 2, the body undergoes an angular displacement

given by 21

The definition of angular displacement holds not only for the rigid body as a whole but also for every particle within the body.The angular displacement

 of a rotating body can be either positive or negative, depending on whether the body is rotating in the direction of increasing (counterclockwise)or decreasing (clockwise).5.Angular velocity

(1).Suppose that our rotating body is at angular position

1

at time t1 and at angular position 2 at time t2.We define the average angular velocity of the body in the time interval t from t1 to t2 to be

21t2t1t

In which t.

is the angular displacement that occurs during

(2).The(instantaneous)angular velocity , with which we shall be most concerned, is the limit of the average angular velocity as t

is made to approach zero.Thus

limdt0tdt

If we know

(t), we can find the angular velocity  by differentiation.(3).The unit of angular velocity is commonly the radian per second(rad/s)or the revolution per second(rev/s).(4).The magnitude of an angular velocity is called the angular speed, which is also represented with .(5).We establish a direction for the vector of the angular velocity  by using rule，a as right-hand shown in the figure.Curl your right hand about the rotating record, your fingers pointing in the direction of rotation.Your extended thumb will then point in the direction of the angular velocity vector.6.Angular acceleration

(1).If the angular velocity of a rotating body is not constant, then the body has an angular acceleration.Let the angular velocity at times

t2

2

and

1

be

and t1, respectively.The average angular acceleration of the rotating body in the interval from t1 to t2 is defined as

In which

21t2t1 t is the change in the angular velocity that occurs during the time interval t.(2).The(instantaneous)angular acceleration , with which we shall be most concerned, is the limit of this quantity as is made to approach zero.Thus

limdt0tdtt

above equations hold not only for the rotating rigid body as a whole but also for every particle of that body.(3).The unit of angular acceleration is commonly the radian per second-squared(rad/s2)or the revolution per second-squared(rev/s2).(4).The angular acceleration also is a vector.Its direction depends on the change of the angular velocity.7.Rotation with constant angular acceleration:

dddt0tdt

d10td(0t)dt0tt2dt2Here we suppose that at time

t0, 00.We also can get a parallel set of equations to those for motion with a constant linear acceleration.8.Relating the linear and angular variables: They have relations as follow: Angular displacement: d Angular velocity：



dsdr

vr Angular acceleration：

atr anv

6.2 Kinetic Energy of Rotation 1.To discuss kinetic energy of a rigid body, we cannot use the familiar formula

Kmv2/2

directly because it applies only to particles.Instead, we shall treat the object as a collection of particles-all with different speeds.We can then add up the kinetic energies of these particles to find kinetic energy of the body as a whole.In this way we obtain, for the kinetic energy of a rotating body，K111122m1v12m2v2m3v3mivi2 2222In which mi is the mass of the ith particle and

vi

is its speed.The sum is taken over all the particles in the body.2.The problem with above equation is that

vi

is not the same for all particles.We solve this problem by substituting for v in the equation with

r, so that we have

211Kmi(ri)2(miri2)22

In which  is the same for all particles.3.The quantity in parentheses on the right side of above equation tells us how the mass of the rotating body is distributed about its axis of rotation.(1).We call that quantity the rotational inertia(or moment of inertia)I of the body with respect to the axis of rotation.It’s a constant for a particular rigid body and for a particular rotation axis.We may now write

Imiri2

(2).The SI unit for I is the kilogram-square meter(kgm2).(3).The rotational inertia of a rotating body depends not only on its mass but also on how that mass is distributed with respect to the rotation axis.4.We can rewrite the kinetic energy for the rotating object as

K1I22

Which gives the kinetic energy of a rigid body in pure rotation.It’s the angular equivalent of the formula

2KMvcm/2, which gives the kinetic energy of a rigid body in pure translation.6.3 Calculating the Rotational Inertia 1.If a rigid body is made up of discrete particles, we can calculate its rotational inertia from

Imiri2.2.If the body is continuous, we can replace the sum in the equation with an integral, and the definition of rotational inertia becomes

Ir2dm.In general, the rotational inertia of any rigid body with respect to a rotation axis depends on(1).The shape of the body,(2).The perpendicular distance from the axis to the body’s center of mass, and(3).The orientation of the body with respect to the axis.The table gives the rotational inertias of several common bodies, about various axes.Note how the distribution of mass relative to the rotational axis affects the value of the rotational inertia I.We would like to give the example of rotational inertia for

a thin circular plate

Irrddr2R0rdr320111dR42(R2)R2mR2

422a thin rod(1)I1r2drl2l213r3l2l21l11()32(l)l2ml2 321212(2)

13l03l1I2I1m()2ml2212I2r2drl1(l)l23

341ml2ml2ml2121238 3.The parallel-axis theorem: If you know the rotational inertia of a body about any axis that passes through its center of mass, you can find its rotational inertia about any other axis parallel to that axis with the parallel-axis theorem:

IIcmMh2

Here M is the mass of the body and h is the perpendicular distance between the two parallel axes.4.Proof of the parallel-axis theorem: Let O be the center of mass of the arbitrarily shaped body shown in cross section in the figure.Place the origin of coordinates at O.Consider an axis

through

O perpendicular to the plane of the figure, and another axis of P parallel to the first axis, Let the coordinates of P be a and b.Let dm be a mass element with coordinates x and y.The rotational inertia of the body about the axis through P is then

Ir2dm[(xa)2(yb)2]dm

Which we can rearrange as I(x2y2)dm2axdm2bydm(a2b2)dmIcm00MhIcmMh22 6.4 Newton’s Second Law for Rotation

1.Torque: The following figure shows a cross section of a body that is free to rotate about an axis passing through O and perpendicular to the cross section.A force F is applied at point P, whose position relative to O is defined by a position vector r.Vector F and r make an angle  with each other.(For simplicity, we consider only forces that have no component parallel to the rotation axis: thus, F is in the plane of the page).We define the torque  as a vector cross product of the position vector and the force

rF 

Discuss the direction and the magnitude of the torque.2.Newton’s second law for rotation(1).The figure shows a simple case of rotation about a fixed axis.The rotating rigid body consists of a single particle of mass m fastened to the end of a massless rod of length r.A force F acts as shown, causing the particle to move in a circle about the axis.The particle has a tangential component of acceleration governed by Newton’s second law: acting on the particle is

Ftmatat

.The torque

.rFFtrmatrm(r)r(mr2)The quantity in parentheses on the right side of above equation is the rotation inertia of the particle about the rotation axis.So the equation can be reduced to

I.(2)For the situation in which more than one force is applied to the particle, we can extend the equation as I.Where  is the net torque(the sum of all external torques)acting on the particle.The above equation is the angular form of Newton’s second law.(3)Although we derive the angular form of Newton’s second law for the special case of a single particle rotating about a fixed axis, it holds for any rigid body rotating about a fixed axis, because any such body can be analyzed as an assembly of single particles.6.5 Work and Rotational Kinetic Energy 1.Work-kinetic energy theorem: Let’s again consider the situation of the figure, in which force F rotates a rigid body consisting of a single particle of mass m fastened to the end of a massless rod.During the rotation, Force F does work on the body.Let us assume that the only energy of the body that changed by F is the kinetic energy.Then we can apply the work-kinetic energy theorem to get

KKfKiWK1212IfIiW22

Above equation is the angular equivalent of the work-kinetic energy theorem for translational motion.We derive it for a rigid body with one particle, but it holds for any rigid body rotated about a fixed axis.2.We next relate the work W done on the body in the figure to the torque  on the body due to force F.If the particle in Fig.11-17 were move a differential distance ds along its circular path, the body would rotate through differential angle with dsrdd,..We would get

dWFdsFtdsFtrddThus the work done during a finite angular displacement from i to f is then

Wd.Above equation holds for

if12 any rigid body rotating about a fixed axis.3.We can find the power P for rotational motion

PdWd dtdt 13